Why does $0,\bar{9}$ equal $1$?

Question

I am finding hard to understand why $0,99999..... = 1$

I have the following proof:

Let $x$ be $0,9999...$

then $10x = 9,999...$

So $10x - x = 9,999 - 0,9999$

$9x = 9 \rightarrow x = 1$

From a philosophical respective, it does seem legit to me that if the decimal form of the number is never ending, then, at infinity and beyond it "tries" to reach 1, so it's limit to infinity equals 1.

My objection is that:

Consider the set $S = \left\{ 0, 1, 2, 3, ..., n \right \}$

What is the possiblity that from the set S we obtain the number $2$?

$P = \frac{P(a)}{P(S)}$ = $ \frac{1}{n}$

so from there we can see that the possiblity is very low, $0.000...1$ , which again we can consider to be $0$ since :

$ \frac{1}{n}$ = $ \frac{1}{\infty}$ = $0$

But again, if we accept that $P = 0$, then there is no possibility that we can select the number 2 from the set $S$, which is false.

Answer 1

Even though your proof is perfectly valid, here is another one; maybe this convinces you more.

The sum of the terms of an infinite geometric series $a_n$ with first term $a$ and ratio $|r|<1$ is $$S_{\infty}=\frac{a}{1-r}$$

Hence, $$0.999\cdots=\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+\cdots=\frac{\frac{9}{10}}{1-\frac{1}{10}}=1$$

Answer 2

It's important to note that $0.\overline{9}$ doesn't have a limit: it's just a single number.

However, we like to define decimals so that they can be used to represent real numbers. And, in particular, we would like $0.\overline{9}$ to be equal to the limit of the sequence

$$ 0.9, 0.99, 0.999, 0.9999, \cdots $$

It is this sequence that "tries" to reach 1, and it is this sequence that we talk about having a limit equal to 1.

But $0.\overline{9}$ is not a sequence: it's just a number.

---

You make a related mistake in your objection: you confuse the single set

$$ \{ 0, 1, 2, \cdots, n \} $$

for various $n$ with the "limiting" set of all integers

$$ \{ 0, 1, 2, \cdots \} $$

It is worth noting, however, that there are serious technical issues with the notion of a uniformly randomly chosen integer.

(Also "probability zero" doesn't mean "impossible". There are significant dangers in trying to use one's intuition about "selecting" randomly from finitely many objects when studying more general probability spaces)

Answer 3

The problem people tend to have with this concept is that we sometimes assume our number system is the exact representation of the numbers themselves. We are, in fact, dealing with a system similar to $\mathbb Z_{10}$, where we roll over to 0 again in a particular position after reaching 9. There isn't a representation for 10 in this system, but instead a 1 in the 10's place followed by a 0 in the 1's place. That said, representing thirds in this system is difficult, so if we change the base from 10 to 3, we can get an exact representation of $0.\overline{9}$ like so:

Given $1_{3} = 1_{10}$ and $0.1_{3} = 0.\overline{3}_{10}$,

\begin{aligned} & 0.1_{3}\\ + & 0.1_{3}\\ \hline & 0.2_{3}\\ and \\ & 0.\overline{3}_{10} \\ + & 0.\overline{3}_{10} \\ \hline & 0.\overline{6}_{10} \\ thus \\ & 0.2_{3}\\ + & 0.1_{3}\\ \hline & 1.0_{3}\\ and \\ & 0.\overline{6}_{10} \\ + & 0.\overline{3}_{10} \\ \hline & 0.\overline{9}_{10} = 1_{3} = 1_{10}\\ \end{aligned}